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Algebraische Geometrie
Universität/Hochschule 
Thema eröffnet von: xiao_shi_tou_
Geometrischer Ansatz zur Collatz-Vermutung?  
Beitrag No.2 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-06-29 17:11
xiao_shi_tou_
 

2020-06-27 17:16 - Slash in Beitrag No. 1 schreibt:
Ich weiß nicht, ob dieses Resultat passend ist für das was du vorhast.



Berg, L., and Meinardus, G., Functional equations connected with the Collatz problem, Results in Math. 25 (1994)

Hi Slash,
da ich ein absoluter Neuling auf diesem Gebiet bin und es auch ehrlichgesagt nur der Vorlesung zur Liebe mache ist das Resultat neu und interessant für mich. Vielen Dank.
Vielleicht melde ich mich hier nochmal, wenn ich Neuigkeiten habe.
MfG XST

Algebraische Geometrie
Universität/Hochschule 
Thema eröffnet von: xiao_shi_tou_
Geometrischer Ansatz zur Collatz-Vermutung?  
Themenstart
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-06-27 16:40
xiao_shi_tou_
 

Hi zusammen.
Im Zuge einer Vorlesung die sich damit beschäftigt Arbeiten über die Collatz-Vermutung zu studieren kam mir die Frage, ob es bisher schon einen geometrischen Ansatz gibt, also mit Hilfe der Algebraischen Geometrie, oder ob das grundsätzlich eher aussichtslos ist.

Ich habe das hier gefunden, was ja sehr interessant ist, aber vielleicht mit dem ursprünglichen Problem nicht mehr viel zu tun hat.

Ich selbst hab das Gefühl, dass ein solcher Ansatz eher nicht funktionieren würde, ich kann mir auch nicht vorstellen wie man das Problem in die Algebraische Geometrie übersetzen könnte, aber im Forum gibt es glaube ich ein paar die sich schon mit dem Problem auseinandergesetzt haben.

Vielleicht seit ihr ja bei der Recherche nach Arbeiten schonmal auf etwas interessantes in diese Richtung gestoßen!?
Ich freue mich auch über Vorschläge zu interessanten Arbeiten über das Thema.
Es geht mir nicht darum das Problem zu lösen, was ich mit 100%ger Wahrscheinlichkeit nicht schaffen würde (:D), sondern darum, dass ich im Zuge der Vorlesung einen Vortrag über eine Arbeit vorbereiten soll. Da würde mich persönlich eben eine Arbeit mit geometrischen Mitteln sehr reizen, aber ich konnte nicht mehr finden als die oben zitierte.

MfG
XST

Ringe
Universität/Hochschule 
Thema eröffnet von: geeert
Idealquotienten und Primärzerlegungen von Idealen  
Beitrag No.1 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-06-21 14:06
xiao_shi_tou_
 

Hi geeert,
schau mal hier.
Grüße
XST
Oder hast du diese Frage auf stackexchange gestellt? :D
Vielleicht steht in Eisenbud ja auch etwas dazu!?

Aktuelles und Interessantes
Universität/Hochschule 
Thema eröffnet von: Mat2
Geschenk für Mathematiker  
Beitrag No.50 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-05-20
xiao_shi_tou_
 
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2020-05-20 22:22 - mibe201067 in Beitrag No. 48 schreibt:
www.amazon.de/HXGL-W%C3%BCrfel-Geschwindigkeits-W%C3%BCrfel-Erwachsener-Berufswettbewerb-Reibungslose/dp/B082ZDD5QR

Das wäre ein schönes Geschenk, aber nur als Anschauungsobjekt. Man sieht den Würfel sonst nie wieder so, wie im Anfangszustand. :-)
Für $1.241,85$ Euro ein bisschen Plastik kaufen ist mir zu teuer. Da spende ich das Geld lieber. Man kann diese Würfel auch mit einem Programm simulieren lassen, das ist kostenlos und reicht erstmal zum üben.
Ich würde als Mathegeschenk einen digitalen Stift empfehlen. Man spart Papier für Notizen und kann sich online besser über Mathe unterhalten.
Kostenpunkt etwa 30-50 Euro.
MfG
\(\endgroup\)

Schwarzes Brett
  
Thema eröffnet von: Tetsuya
Serlo Informatik: Teile deine Leidenschaft mit anderen!  
Beitrag No.8 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-05-16
xiao_shi_tou_
 

@Tetsuya
Ob Serlo jetzt wirklich besser erklärt als Wikipedia sehe ich jetzt nicht sofort ein.
Fest steht, dass man bei Serlo aber schnell an Grenzen stößt. Wirklich viele Inhalte gibt es dort noch nicht, aber die Seite befindet sich ja auch noch im Aufbau.
Der Matheplanet ist hingegen einzigartig, hier gibt es sowohl interessante Artikel als auch eine Community die bei Fragen gezielt weiterhilft.
VG
XST

Schwarzes Brett
  
Thema eröffnet von: Tetsuya
Serlo Informatik: Teile deine Leidenschaft mit anderen!  
Beitrag No.4 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-05-15
xiao_shi_tou_
 

Analysis Eins:
"Stell Dir vor, hochwertige Bildung steht welt-weit allen Menschen komplett kostenlos zur Verfügung und wird von Kindern, SchülerInnen und Studierenden aktiv mitgestaltet."

Das gibt es doch schon lange in Form von Wikipedia.
Was genau ist also der Bewegungsgrund extra noch eine neue Platform aufzumachen?
Bisher sehe ich keinen Grund Serlo Wikipedia vorzuziehen:
Es gibt weniger Inhalte und man weiß auch nicht inwiefern die Inhalte auf Serlo korrekt sind (was man bei Wikipedia leider auch nicht immer weiß, aber dort schauen mehr Experten drauf würde ich mal vermuten).

Viele Grüße
XST

PS:Ich fühle mich hier auf dem MP wohl.

Relationen und Abbildungen
Universität/Hochschule 
Thema eröffnet von: Tres194
Relationen  
Beitrag No.8 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-05-13
xiao_shi_tou_
 
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2020-05-13 20:13 - helmetzer in Beitrag No. 4 schreibt:
2020-05-13 19:22 - xiao_shi_tou_ in Beitrag No. 3 schreibt:
Beachte auch, dass eine asymmetrische Relation nicht symmetrisch ist. Kann also eine symmetrische Relation asymmetrisch sein?
Auch wenn es an Haarspalterei grenzt: Bei einer leeren Relation sollte man das noch einmal überdenken.

Uups. Stimmt, so einfach ist es nicht.
Naja, dann muss man benützen, dass die Implikation $(a,b)\in R\implies (b,a)\not\in R$ immer gilt, da es keine $(a,b)\in R$ gibt.
Naja, Haarspalterei ist es ja nicht, wenn nur "wahr" und "nicht wahr" als Lösungen in Frage kommen und ich meine Bemerkung genau die falsche Lösung impliziert.
Gut, dass es dir aufgefallen ist.
VG XST
\(\endgroup\)

Relationen und Abbildungen
Universität/Hochschule 
Thema eröffnet von: Tres194
Relationen  
Beitrag No.3 im Thread
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xiao_shi_tou_
 

Hi
Die leere Relation kann (auf einer nicht-leeren Menge wie den natürlichen Zahlen) nicht reflexiv sein, da es ja nicht einmal ein einziges Element gibt welches in Relation zu sich selbst steht, geschweige denn alle.
Beachte auch, dass eine asymmetrische Relation nicht symmetrisch ist. Kann also eine symmetrische Relation asymmetrisch sein?

Relationen und Abbildungen
Universität/Hochschule 
Thema eröffnet von: MePep
Gröbste Verfeinerung - Äquivalenzrelationen  
Beitrag No.7 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-05-09
xiao_shi_tou_
J
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2020-05-09 13:52 - MePep in Beitrag No. 6 schreibt:
Moment, ist also \(\sim\) die Schnittmenge von \(\sim_{1} und \sim_{2}\) ?
Was sollte es sonst sein? Es gibt keine größere (also gröbere) Menge die sowohl in $R_{\sim_1}$ als auch in $R_{\sim_2}$ enthalten ist.
Du solltest noch nachschauen, dass die Schnittmenge eine Äquivalenzrelation definiert.
Und jetzt, wo du weißt wie die gröbste Verfeinerung definiert wird musst du auch nicht mehr den Umweg über die Teilmengen $R$ gehen, sondern kannst die Definition der gröbsten Verfeinerung direkt hinschreiben und alles nachrechnen.
Viel Erfolg
\(\endgroup\)

Relationen und Abbildungen
Universität/Hochschule 
Thema eröffnet von: MePep
Gröbste Verfeinerung - Äquivalenzrelationen  
Beitrag No.4 im Thread
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xiao_shi_tou_
J
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Hi.
Wenn $\sim$ eine Äquivalenzrelation auf einer Menge $X$ ist, dann kannst du eine Menge $R_{\sim}\sube X\tm X$ definieren  durch
$(x,x')\in R_{\sim}$ genau dann wenn $x\sim x'$.
Es ist nicht schwer sich klarzumachen welche Eigenschaften diese Menge $R_\sim$ haben muss. Zum Beispiel muss sie die Diagonale enthalten, wegen Reflexivität.

Eine Äquivalenzrelation $\sim$ ist feiner als eine Äquivalenzrelation $\sim'$ genau dann wenn aus $x\sim x'$ immer $x\sim' x'$ folgt, also wenn $R_{\sim}\sube R_{\sim'}$ gilt.
Eine gemeinsame Verfeinerung von $\sim_1,\sim_2$ finden bedeutet also eine Menge $R\sube X\times X$ zu finden die sowohl in $R_{\sim_1}$ als auch in $R_{\sim_2}$ enthalten ist und eine Äquivalenzrelation definiert. Es ist klar, was es bedeutet die gröbste solche Menge zu finden. Welcher Kandidat kommt in Frage?
\(\endgroup\)

Moduln
Universität/Hochschule 
Thema eröffnet von: Stocksn
Kurze exakte Sequenzen und direkte Summe  
Beitrag No.1 im Thread
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xiao_shi_tou_
 
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Gegeben ist also eine kurze exakte Folge
$$0\to M\overset{f}{\to} N\overset{g}{\to} P\to 0$$ von $R$-Moduln.
Das ist schonmal die Voraussetzung.
Leider ist der Post abgeschnitten.
Was möchtest du zeigen/fragen?
MfG XST
\(\endgroup\)

Holomorphie
Universität/Hochschule 
Thema eröffnet von: Drumbene91
Cauchy-Riemann in Polarkoordinaten  
Beitrag No.1 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-04-22
xiao_shi_tou_
J
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Sei $f(z)=u(z)+iv(z)$.
Dann ist $f$ genau dann komplex diffbar, wenn die Cauchy-Riemannschen Gleichungen
$$\partial_x u=\partial_y v, \partial_y u=-\partial_x v$$ gelten.
Die Bedingung in der Aufgabe
ist offenbar äquivalent zu
$$\partial_\theta u+i\partial_\theta v=i\partial_t u-\partial_t v,$$ also zu
$$\partial_\theta u=-\partial_t v, \partial_\theta v=\partial_t u.$$ Der Zusammenhang zwischen den $\partial_x u,\partial_y u,\partial_x v,\partial_y v$ und den $\partial_\theta u,\partial_\theta v,\partial_t u,\partial_t v$ ist durch die Kettenregel gegeben, zum Beispiel gilt:
$\partial_\theta u(x,y)=\partial_x u\partial_\theta x+\partial_y u\partial_\theta y.$
Viele Grüße
XST
\(\endgroup\)

Funktionen
Universität/Hochschule 
Thema eröffnet von: hari01071983
x^2 ist (nicht) injektiv  
Beitrag No.15 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-04-22
xiao_shi_tou_
 
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Ist es dir überlassen die Definitionsbereiche zu wählen?
Beachte, dass man ganz einfache Beispiele von injektiven Funktionen mit $x\mapsto x^2$ bekommt, wenn man zum Beispiel als Definitionsbereich $\emptyset$ oder eine Einpunktmenge nimmt, denn dann gibt es gar nicht $2$ Punkte die auf ein und denselben Punkt abgebildet werden könnten 🙂.
Ansonsten kannst du so vorgehen wie oben beschrieben.
\(\endgroup\)

Zahlentheorie
Schule 
Thema eröffnet von: ziad38
Ist es möglich, diesen Beweis zu verstehen?  
Beitrag No.16 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-04-05
xiao_shi_tou_
 

2020-04-04 16:25 - Nuramon in Beitrag No. 4 schreibt:
2020-04-04 16:09 - ziad38 in Beitrag No. 2 schreibt:
wooow Du hast Recht, warum habe ich nicht auch so gedacht obwohl ich 3  mal gelesen habe.
Ich habe mich auch schon mal über ein [1] am Ende einer Formel in einem Artikel gewundert, bis mir der Betreuer meiner Bachelorarbeit sagte, dass das eine Quellenangabe ist.


den beweis noch ?
Welche Fragen hast du zu dem Beweis?

[Die Antwort wurde nach Beitrag No.2 begonnen.]
Naja, das ist ja noch halb so schlimm.
Ich hab Ewigkeiten versucht eine Aussage zu beweisen, die eigentlich als Definition gedacht war...Dennoch hat sich daraus jetzt etwas interessantes ergeben.

[Die Antwort wurde nach Beitrag No.11 begonnen.]

Teilbarkeit
  
Thema eröffnet von: OliverFuchs
ggT(a², b²) = ggT(a, b)²  
Beitrag No.16 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-03-25
xiao_shi_tou_
 
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2020-03-25 16:33 - Nuramon in Beitrag No. 14 schreibt:
2020-03-25 16:25 - xiao_shi_tou_ in Beitrag No. 13 schreibt:
Auf deinem Profil gibt es ein Latex Profil.
Da kannst du Abkuerzungen wie
\newcommand{\Q}{\mathbb{Q}} 
hinschreiben. Aber dort keine Dollarzeichen verwenden.
Dann musst du nur noch
$\Q$
schreiben, wenn du $\Q$ haben willst.

Bei default gibt es auf dem MP schon \IQ, \IR, \IN, \IZ, \IC für $\IQ, \IR, \IN, \IZ, \IC$. Wenn man sich den einen extra Buchstaben leisten kann, dann braucht man muss man sich also dafr gar nicht extra Abkürzungen einrichten.
Gut zu wissen, wusst ich nicht👍😎.
\(\endgroup\)

Teilbarkeit
  
Thema eröffnet von: OliverFuchs
ggT(a², b²) = ggT(a, b)²  
Beitrag No.13 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-03-25
xiao_shi_tou_
 
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\newcommand{\Teee}{[T_1,T_2,T_3]} \newcommand{\Ten}{[T_1\cos T_n]} \newcommand{\Tem}{[T_1\cos T_m]} \newcommand{\pts}{\dotsc} \newcommand{\pt}{\cdot} \newcommand{\hm}[3]{\Hom_{#1}(#2,#3)} \newcommand{\hom}{\Hom} \newcommand{\dash}{\dashrightarrow} \newcommand{\schemes}{\bb{(Sch)}} \newcommand{\groups}{\bb{(Grp)}} \newcommand{\rings}{\bb{(Ring)}} \newcommand{\tx}[1]{\text{ #1 }} \newcommand{\mm}{\ff{m}} \newcommand{\zkinfsum}{\sum_{k=0}^\infty} \newcommand{\ziinfsum}{\sum_{i=0}^\infty} \newcommand{\zjinfsum}{\sum_{j=0}^\infty} \newcommand{\asum}[1]{\sum_{\a\in\N^n}#1 X^\a} \newcommand{\arr}[3]{#1\overset{#2}{\to} #3} \newcommand{\nrm}[1]{\left\|#1\right\|} \newcommand{\nr}{\nrm{-}} \newcommand{\ext}[2]{#1/{#2}} \newcommand{\lam}{\lambda} \newcommand{\a}{\alpha} \newcommand{\be}{\beta} \newcommand{\gam}{\gamma} \newcommand{\de}{\delta} \newcommand{\vp}{\varphi} \newcommand{\p}{\phi} \newcommand{\bul}{\bullet} \newcommand{\t}{\tau} \newcommand{\s}{\sigma} \newcommand{\ze}{\zeta} \newcommand{\T}{\mathbb{T}} \newcommand{\tm}{\times} \newcommand{\tms}{\times\pts\times} \newcommand{\ot}{\otimes} \newcommand{\ots}{\otimes\pts\otimes} \newcommand{\pls}{+\pts +} \newcommand{\cos}{,\pts,} \newcommand{\op}{\oplus} \newcommand{\ops}{\oplus\pts\oplus} \newcommand{\cr}{\circ} \newcommand{\crs}{\circ\pts\circ} \newcommand{\sc}[1]{\mathscr{#1}} \newcommand{\scal}[2]{\sc{#1}{\!#2}} \newcommand{\ov}[2]{\begin{matrix}#1 \\ #2\end{matrix}} \newcommand{\viele}{\color{orange}{\udl{\color{black}{\sc{V}\!iele\tx{}\sc{G}\!r\overset{{}_{,,\!}}{u}\textit{ß}e}}}} \newcommand{\xst}{\color{orange}{\udl{\color{black}{X.S.T.\sim 小石头}}}} \newcommand{\gudl}[1]{\color{orange}{\udl{\color{black}{#1}}}} \newcommand{\Task}{\gudl{\sc{T}\!ask:}} \newcommand{\Exer}{\gudl{\sc{E}\!exercise:}} \newcommand{\Drinfeld}{\gudl{\sc{D}\!rinfeld:}} \newcommand{\Goss}{\gudl{\sc{G}\!oss}} \newcommand{\CK}{C/K} \newcommand{\CS}{C/S} \newcommand{\Ck}{C/k} \newcommand{\Om}{\Omega} \newcommand{\J}{\Jac_{\CS}^{g-1}} 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Hi.
In einem Ring nennt man multiplikative Inverse Einheiten.
Im Fall von $\Z$ sind das $\pm 1$.
Man bekommt im allgemeinen in einem sogenannten UFD (Ein Ring mit eindeutiger Primfaktorzerlegung) eine Zerlegung die eindeutig ist bis auf Multiplikation mit Einheiten.
Das kannst du alles nachlesen unter dem Stichwort UFD oder ZPE-Ring.

TeX-code kann man auch mit Dollarzeichen machen:
TeX
$\mathbb{R}$
$\mathrm{ggT}$

$\mathbb{R}$
$\mathrm{ggT}$



[Die Antwort wurde nach Beitrag No.10 begonnen.]

PS: Auf deinem Profil gibt es ein Latex Profil.
Da kannst du Abkuerzungen wie
\newcommand{\Q}{\mathbb{Q}} 
hinschreiben. Aber dort keine Dollarzeichen verwenden.
Dann musst du nur noch
$\Q$
schreiben, wenn du $\Q$ haben willst.
\(\endgroup\)

Teilbarkeit
  
Thema eröffnet von: OliverFuchs
ggT(a², b²) = ggT(a, b)²  
Beitrag No.8 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-03-25
xiao_shi_tou_
 
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2020-03-25 15:36 - OliverFuchs in Beitrag No. 5 schreibt:
Ja danke nochmal. Zum einen für die Willommensgrüße im Forum zum anderen
für die schnelle Hilfe. Das mit der Linearkombination von a und b,
hätte ich auch sehen können. Sorry
Aber das was der Kollege mit p/ggT(a^2,b^2) gesagt hat, da hat er recht.
Genau so habe ich es auch in meiner Version gemacht nur habe ich es
aus dem Grund nicht verwendet, weil ich mir nicht sicher war ob man
bei der Verwendung der Primzahl nicht Argumentte mit einbaut,
mit denen man auch die Primfaktorzerlegung zeigen kann. Da habe ich
zu wenig parat wie man diese zeigt und welch Argumente da eingehen.
Wenn ich mich aber recht aus der Vorlesung erinnere, so kann man die Primfaktorzerlegung in jedem Ring zeigen, in welchem auch die Divison mit
Rest gilt. Nur muss sie dann nicht eindeutig sein. Dazu, glaube ich wenigstens, benötigt man den Hauptsatz der Arithmetik (p/a1...an =>
p/ai 1<=i<=n). Wenn also bei der alternativen Beweisführung nicht beide
Dinge verwendet werden dann wäre der Beweis zulässig. Das muss
ich mir noch ansehen.
Danke nochmals für die Hilfe und den Willkommensgruß
lg Oliver
Aus algebraischer Sicht ist ja die PFZ in dem Ring $\Z$ auch nicht eindeutig, da man die Wahl zwischen $\pm p_i$ hat. Man legt sich nur traditionell auf $p_i>0$ fest.
Dass in jedem Ring mit Euklidischer Division es eine Primfaktorzerlegung gibt ist richtig.

Ich muss mich jetzt doch noch korrigieren
Man braucht die Primfaktorzerlegung schon. Tut mir leid für diese falsche Aussage.

[Die Antwort wurde nach Beitrag No.6 begonnen.]
\(\endgroup\)

Teilbarkeit
  
Thema eröffnet von: OliverFuchs
ggT(a², b²) = ggT(a, b)²  
Beitrag No.3 im Thread
Zum letzten BeitragZum nächsten BeitragZum vorigen BeitragZum erstem Beitrag2020-03-25
xiao_shi_tou_
 
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$\newcommand{\ggT}{\mathrm{ggT}}$
Zeige $\ggT(a^2,b^2)=\ggT(a,b)^2$ ohne Verwendung der Primfaktorzerlegung.

Ich habe mir einen Beweis gebastelt, wo ich mir aber unsicher bin.
Ich glaube der taugt auch nicht viel. Dann hat der Professor die folgenden
Lösung veröffentlicht.

Diese Aufgabe soll zeigen, dass man manche Eigenschaften des $\ggT$ nicht alleine aus der Definition folgern kann.  Um $\ggT(a^2,b^2) =\ggT(a,b)^2$ zu zeigen,  muss man auch die Theorie der linearen diophantischen Gleichungen verwenden.

Sei wieder $d:=\ggT(a,b)$.  Aus  $d|a$ und $d|b$ folgt $d^2|a^2$ und $d^2|b^2$, sodass $\underline{d^2|\ggT(a^2,b^2)}$. $\color{green}{\checkmark}$

Umgekehrt müssen wir auch zeigen, dass $\ggT(a^2,b^2)|d^2$, und das kann man nicht direkt aus  der  Definition  des  $\ggT$  ableiten.  

Warum nicht?
Wir müssen nur für alle Primzahlen $p$ zeigen:
$p\mid \ggT(a^2,b^2)\implies p\mid d^2$.
Das ist trivialerweise der Fall aufgrund der Definition einer Primzahl. Es wurde nicht die Primfaktorzerlegung benutzt.

Darum  zeigen  wir  es,  indem  wir  nachweisen,  dass $a^2x+b^2y=d^2$ eine  ganzzahlige  Lösung  $(x,y)$  hat,  zumal  in  der  Vorlesung  bewiesen wurde, dass alle Zahlen, die sich als Linearkombination von $a^2$ und $b^2$ darstellen lassen,Vielfache von $\ggT(a^2,b^2)$ sind.

Aus der Definition von d folgt die Existenz ganzer $u,v$ mit $au+bv=d$.  Durch Potenzierenmit $3$ folgt $(au+bv)^3=a^3u^3+ 3a^2u^2bv+ 3aub^2v^2+b^3v^3=a^2(au^3+ 3u^2bv) +b^2(3auv^2+bv^3) =d^3$.

Nun sind beide Klammerausdrücke durch $d$ teilbar, sodass mit $x= (au^3+ 3u^2bv)/d$ und
$y=(3auv^2+bv^3)/d$ eine ganzzahlige (!)  Lösung von $a^2x+b^2y=d^2$ gefunden ist. $\color{green}{\checkmark}$

(Ja, so kann mans auch machen.)

Was nun ich nicht sehen kann ist, warum $d$ die beiden Klammernausdrücke
teilen soll. Wenn mir den ersten ansehe und $v^2$ heraushebe, erhalte ich
$v^2(3au +bv)$. Nun ist $d=au+bv$. Wie sieht man das die Teilbarkeit?

Antwort
Das ist deshalb so, weil alle Zahlen der Form $aX+bY$ durch $d$ teilbar sind, wie du selbst schon gesagt hast (... in der Vorlesung bewiesen wurde ). Beide Klammern haben diese Form für gewisse $X,Y$

lg Oliver Fuchs

[Die Antwort wurde vor Beitrag No.1 begonnen.]

EDIT: Phillippw hat es eigentlich schon auf den Punkt gebracht 🙃
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Beitrag No.7 im Thread
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xiao_shi_tou_
 

Hast du gut gemacht👌👍.

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xiao_shi_tou_
 

2020-03-11 15:01 - MontyPythagoras in Beitrag No. 14 schreibt:
2020-03-11 14:21 - xiao_shi_tou_ in Beitrag No. 13 schreibt:
Ich wäre jetzt von einem Idealfall ausgegangen in denen alle Zahlen stimmen. Welche Methode theoretisch, bei korrekten Zahlen die genaueste ist, das ist mein Anliegen. Dass es zum Beispiel eine hohe Dunkelziffer gibt ist ja durch aus ein realistischer Faktor der aber mit der Methode an sich nichts zu tun hat, sondern nur mit dem Input.

Würdest Du Dich auch mit der Frage auseinandersetzen, ob Du den Ausgang eines Pferderennens exakt vorhersagen kannst, wenn Du nur alle Input-Daten genau kennst? Muskelmasse und Körperfett jedes Rennpferds, Alter des Pferdes, Beinlängen, Vo2max-Werte, aerobe und anaerobe Schwelle, Kaloriengehalt und Zusammensetzung des letzten Futters, Gewicht des Jockeys (Ballast) usw., wenn es doch völlig aussichtslos ist, diese Zahlen jemals in der erforderlichen Güte zu bekommen? Gut, ein Mathematiker sagt jetzt wahrscheinlich "ja". 😛

Ciao,

Thomas
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